Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $t = \dfrac{z - 10}{-9z^2 - 63z} \div \dfrac{7z + 21}{z^2 + 10z + 21} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{z - 10}{-9z^2 - 63z} \times \dfrac{z^2 + 10z + 21}{7z + 21} $ First factor the quadratic. $t = \dfrac{z - 10}{-9z^2 - 63z} \times \dfrac{(z + 7)(z + 3)}{7z + 21} $ Then factor out any other terms. $t = \dfrac{z - 10}{-9z(z + 7)} \times \dfrac{(z + 7)(z + 3)}{7(z + 3)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ (z - 10) \times (z + 7)(z + 3) } { -9z(z + 7) \times 7(z + 3) } $ $t = \dfrac{ (z - 10)(z + 7)(z + 3)}{ -63z(z + 7)(z + 3)} $ Notice that $(z + 3)$ and $(z + 7)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ (z - 10)\cancel{(z + 7)}(z + 3)}{ -63z\cancel{(z + 7)}(z + 3)} $ We are dividing by $z + 7$ , so $z + 7 \neq 0$ Therefore, $z \neq -7$ $t = \dfrac{ (z - 10)\cancel{(z + 7)}\cancel{(z + 3)}}{ -63z\cancel{(z + 7)}\cancel{(z + 3)}} $ We are dividing by $z + 3$ , so $z + 3 \neq 0$ Therefore, $z \neq -3$ $t = \dfrac{z - 10}{-63z} $ $t = \dfrac{-(z - 10)}{63z} ; \space z \neq -7 ; \space z \neq -3 $